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The polynomial $X^4+5X^3-2X^2-2$ has no roots in $\Bbb{Z}_{11}$ so I am unsure as to how I am meant to factorise in such a scenario when I cannot use the factor theorem. How am I meant to progress? And in general for any polynomial with no roots in some finite field $\Bbb{Z}_p$ how should I go about telling if it is already irreducible, and if it isn't how do I know how to factor it?

Bill Dubuque
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awgya
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    If a degree $4$ polynomial factors and has no linear factors, what can you conclude? – Robert Shore May 04 '22 at 22:20
  • You can follow the Cantor-Zassenhaus algorithm. Square-and-multiply is your friend in calculating the remainders of high powers, and you "quickly" see that your polynomial $f(x)$ is a factor of $x^{121-1}-1$ meaning that it, indeed, is a product of two irreducible quadratics. When we calculate $\gcd(f(x),x^{60}-1)$ we are unlucky. That gcd is equal to $1$ meaning that all the zeros (in $\Bbb{F}_{121}$ are non-squares. We have better luck with $\gcd(f(x),(x-1)^{60}-1)=x^2-2x-1$. This gcd finds the zeros of $f(x)$ that differ from a square by $1$. – Jyrki Lahtonen May 05 '22 at 03:00
  • Of course, your field is relatively small, and you can do trial and error as in José's writings. Possibly (likely?) that is faster in your case. I don't have a clear idea of where the break even -point is :-). For more about Cantor-Zassenhaus, see here and the links in there. – Jyrki Lahtonen May 05 '22 at 03:04

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If $X^4+5X^3-2X^2-2$ has no first degree factors, then it is reducible if and only if it can be written as the product of two quadratic polynomials $X^2+aX+b$ and $X^2+cX+d$. If they exist, then$$\left(X^2+aX+b\right)\left(X^2+cX+d\right)=X^4+5X^3-2X^2-2,$$and therefore $a+c=5$ and $bd=-2=9$. So, for each $b\in\Bbb Z_{11}\setminus\{0\}$, take $d\in\Bbb Z_{11}$ such that $bd=9$, and then see whether or not$$\left(X^2+aX+b\right)\left(X^2+(5-a)X+d\right)=X^4+5X^3-2X^2-2.$$It turns out that, if you pick $b=-1$ and $d=2$, then $a=-2$ will do. That is,$$\left(X^2-2X-1\right)\left(X^2+7x+2\right)=X^4+5X^3-2X^2-2.$$

José Carlos Santos
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  • I understand the method you have used but how have you chosen $b=-1$ and $d=2$? – awgya May 05 '22 at 12:27
  • Trial and error. I first took $b=1$ and $d=-2$, but then no $a$ will work. And then I took $b=-1$ and $d=2$, and all went fine. – José Carlos Santos May 05 '22 at 12:43
  • Forgive me if what I am asking is obvious but if it's just trial and error what about all the other numbers in $\Bbb{Z}_{11}$ that make $9$, like $3\times 3$ and $4\times 5$, etc. Do I have to keep trying numbers until some work or is there some common sense which says what values it's more likely to be? – awgya May 06 '22 at 11:27
  • I just kept on trying. When I found two numbers $b$ and $d$ whose product was $9$ and for which there was a solution, I stopped trying. That's all. No, I had no reason to suppose that $b=-1$ and $d=2$ was more likely to work than $b=d=3$. – José Carlos Santos May 06 '22 at 11:31